Numerical methods for finding the roots of $f(x)=\left(\cos{\frac{33}{x}\pi}\right) (\cos{x \pi})-1$ I have a trigonometric function; for instance
$$f(x)=\left(\cos{\frac{33}{x}\pi}\right) (\cos{x \pi})-1$$ 
I wanted to know the zeroes of this particular function, so I thought I could look into some root-finding algorithms (Newton's, Halley's, Secant...). However, they don't seem to work as $f'(x)=0$ at the roots of $f(x)$, so all those methods are not guaranteed to converge.
So, I was thinking, is there some type of root-finding algorithm for this particular trigonometric equation? Or at least transform this equation into one that the roots would go through the x-axis rather than "bouncing" off it, so Newton's method would apply. 
Also, I am focused on roots $>1$ and $<33$. 

Note: Although the given example can be solved with trigonometric techniques, I am specifically looking for numerical methods. The example was chosen to make it easy to check the roots. I can generalize it to say for any $$f(x)=\left(\cos{\frac{n}{x}\pi}\right) (\cos{x \pi})-1$$ and an interval $$[a,b]$$ where there is only one root in that interval, is there a way to use numerical methods that are guaranteed to converge at the root to find that root?

 A: We have $$\frac{33\pi}{x}=2\pi k,$$ where $k\in\mathbb Z$ and
$$x\pi=2\pi n,$$ where $n\in\mathbb Z$.
We obtain: $$33=4kn,$$ which is impossible.
Also, there is a case 
$$\cos\frac{33\pi}{x}=\cos{\pi x}=-1.$$
Here we obtain:
$$33=(1+2k)(1+2n).$$
Can you end it now?
A: Your task is equivalent to solving $\cos{\frac{33}{x}\pi}=\cos{x \pi}=1$ or  $\cos{\frac{33}{x}\pi}=\cos{x \pi}=-1$. The first equation results in $\frac{33}{x}\pi=2\pi n$, $x=\frac{33}{2n}$ -not a solution because $\cos \frac{33}{2n}\pi \ne 1, n \in Z, n \ne 0$. 
The second equation results in $\frac{33}{x}\pi=\pi(1+2n)$, $x=\frac{33}{2n+1}$. Now we have $\cos \frac{33}{2n+1}\pi =-1$, or $\frac{33}{2n+1}\pi=(2k+1)\pi$, $k \in Z$. We can rewrite the last equation as $(2n+1)(2k+1)=33$ which gives us solutions $(3,11)$, $(-3,-11)$, $(1,33)$, $(-1,-33)$.
As for using numerical methods, there may be difficulty with using Newton method because the function and it's derivative have a lot of points of discontinuity and derivative may have a point of discontinuity where the value of function is zero.
A: For $\cos(x)\cos(y)$ to be equal to $1$, either both $\cos(x)$ and $\cos(y)$ must be equal to $1$ or both equal to $-1$. This is because the range of $\cos(x)$ is $[-1, 1]$. This means we want to solve $$\cos(x\pi) = 1, \cos\left(\frac{33}{x}\pi\right) = 1$$
and $$\cos(x\pi) = -1, \cos\left(\frac{33}{x}\pi\right) = -1$$
Tackling the first case first, for $\cos(t)$ to equal $1$, $t$ must be $2\pi k$, with $k$ an integer. This means $x = 2k_1$ is an integer and $x = \frac{33}{2k_2}$ is an integer. This cannot happen as $33$ has no even divisors.
For the second case, for $\cos(t)$ to equal $-1$, $t$ must equal $\pi + 2\pi k$. This means $x = 1+2k_1$ and $\frac{33}{x} = 1 + 2k_2$. For $33/x$ to be an integer, $x$ must be equal to $\pm 1, \pm 3, \pm 11, \pm 33$. $x$ and $33/x$ for all of these $x$ is odd.
Therefore, the solutions are $x = \pm 1, \pm 3, \pm 11, \pm 33$.
A: In a general manner, if you want to find the zero of $f(x)=0$ knowing that the solution is such that $a < x <b$, a good algorithm is used in subroutine $\color{red}{\text{rtsafe}}$ from Numerical Recipes (have a look here for the source code in C).
Basically, what it does is to combine  bisection steps (whenever Newton method would make the iterate to be out of the given bounds - these are permanently updated) and Newton steps.
